Delta function integral

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  • #1
nicksauce
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Evaluate:

[tex]\int^{\infty}_{-\infty} f(x)\delta(x-x_0)dx[/tex]

Where

[tex]f(x)=ln(x+3), x_0=-2[/tex]

Ordinarily, you would just evaluate [tex]f(x_0)[/tex], so it would be 0, but in this case, since [tex]f(x)[/tex] is [tex]-\infty[/tex] at [tex]x=-3[/tex], does that make a difference?
 
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  • #2
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But you have x0=-2 and f(x) is certainly not negative infinity there. What is f(-2)?
 
  • #3
nicksauce
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Right, f(-2) = ln(1) = 0

I'm just wondering if for the formula:

[tex]\int_{-\infty}^{\infty}f(x)\delta (x-x_0) = f(x_0) [/tex]

there need to be certain conditions on f(x), that would make a logarithmic function not suitable? My textbook was very vague in this area.
 
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  • #4
Hurkyl
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You're missing the + and the - on your limits...


There are ways to define integration of generalized functions so that this integral makes sense for any (partial) function f that is well-behaved near x0. It might even be possible to make this integral well-defined when f is a generalized function that is well-behaved near x0!
 
  • #5
HallsofIvy
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Right, f(-2) = ln(1) = 0

I'm just wondering if for the formula:

[tex]\int_{-\infty}^{\infty}f(x)\delta (x-x_0) = f(x_0) [/tex]

there need to be certain conditions on f(x), that would make a logarithmic function not suitable? My textbook was very vague in this area.
No. For any function defined at x0,
[tex]\int_{-\infty}^\infty f(x)\delta (x)dx= f(x_0)[/tex]
 
  • #6
Hurkyl
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No. For any function defined at x0,
[tex]\int_{-\infty}^\infty f(x)\delta (x)dx= f(x_0)[/tex]
I think I won't feel comfortable unless you insist on continuity in a neighborhood of x0.
 

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